To assess Mendel's law of segregation using tomatoes, a true- breeding tall variety (SS) is crossed with a true-breeding short variety \((s s) .\) The heterozygous tall plants \((S s)\) were crossed to produce the two sets of \(\mathrm{F}_{2}\) data as follows: $$\begin{array}{cc} \text { Set I } & \text { Set II } \\ 30 \text { tall } & 300 \text { tall } \\ 5 \text { short } & 50 \text { short } \end{array}$$ (a) Using chi-square analysis, analyze the results for both datasets. Calculate \(\chi^{2}\) values, and estimate the \(p\) values in both cases. (b) From the analysis in part (a), what can you conclude about the importance of generating large datasets in experimental settings?

Short Answer

Expert verified
Answer: Large datasets are important in experimental settings as they provide stronger statistical evidence. In the case of smaller dataset (Set I), we could not reject Mendel's law of segregation while with the larger dataset (Set II), we had strong evidence to reject it. Poor results from smaller datasets can lead researchers to incorrect conclusions, thus underlining the importance of generating large datasets in experimental settings.

Step by step solution

01

Writing Down Observed Frequencies

Firstly, we need to write down the observed frequencies (O) from the data provided for set I and set II. Set I: Tall plants observed frequency (O_tall Set I) = 30 Short plants observed frequency (O_short Set I) = 5 Set II: Tall plants observed frequency (O_tall Set II) = 300 Short plants observed frequency (O_short Set II) = 50
02

Calculate Expected Frequencies

Now, we need to calculate the expected frequencies (E) for both datasets. The expected ratio of tall to short plants according to Mendel's law of segregation is 3:1. Total plants in Set I = 30 + 5 = 35 Expected frequencies in Set I: E_tall Set I = (3/4) * 35 = 26.25 E_short Set I = (1/4) * 35 = 8.75 Total plants in Set II = 300 + 50 = 350 Expected frequencies in Set II: E_tall Set II = (3/4) * 350 = 262.50 E_short Set II = (1/4) * 350 = 87.50
03

Calculate Chi-square Values

Next, we calculate the Chi-square values for both datasets using the formula: \(\chi^2 = \Sigma\frac{(O-E)^2}{E}\) Set I: \(\chi^2_{Set I} = \frac{(30-26.25)^2}{26.25} + \frac{(5-8.75)^2}{8.75} \approx 0.535 + 1.607 = 2.142\) Set II: \(\chi^2_{Set II} = \frac{(300-262.50)^2}{262.50} + \frac{(50-87.50)^2}{87.50} \approx 5.350 + 20.286 = 25.636\)
04

Estimating P-values

Now, we estimate the p-values using the calculated Chi-square values and 1 degree of freedom (df): Set I: p-value for \(\chi^2_{Set I} = 2.142\) with 1 df is approximately 0.15. Set II: p-value for \(\chi^2_{Set II} = 25.636\) with 1 df is approximately <0.001.
05

Conclusions

(a) For Set I, the p-value is 0.15 which indicates that there is not enough evidence to reject Mendel's law of segregation. For Set II, the p-value is <0.001 which indicates that there is strong evidence to reject Mendel's law of segregation. (b) From the analysis in part (a), we can conclude that large datasets are important in experimental settings as they provide stronger statistical evidence. In the case of smaller dataset (Set I), we could not reject Mendel's law of segregation while with the larger dataset (Set II), we had strong evidence to reject it. Poor results from smaller datasets can lead researchers to incorrect conclusions, thus underlining the importance of generating large datasets in experimental settings.

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Most popular questions from this chapter

Albinism in humans is inherited as a simple recessive trait. Determine the genotypes of the parents and offspring for the following families. When two alternative genotypes are possible, list both. (a) Two parents without albinism have five children, four without albinism and one with albinism. (b) A male without albinism and a female with albinism have six children, all without albinism.

Why was the garden pea a good choice as an experimental organism in Mendel's work?

In Drosophila, gray body color is dominant over ebony body color, while long wings are dominant over vestigial wings. Work the following crosses through the \(F_{2}\) generation, and determine the genotypic and phenotypic ratios for each generation. Assume that the \(P_{1}\) individuals are homozygous: (a) gray, long \(\times\) ebony, vestigial, and (b) gray, vestigial \(\times\) ebony, long, and (c) gray, long \(\times\) gray, vestigial.

Two true-breeding pea plants are crossed. One parent is round, terminal, violet, constricted, while the other expresses the contrasting phenotypes of wrinkled, axial, white, full. The four pairs of contrasting traits are controlled by four genes, each located on a separate chromosome. In the \(F_{1}\) generation, only round, axial, violet, and full are expressed. In the \(\mathrm{F}_{2}\) generation, all possible combinations of these traits are expressed in ratios consistent with Mendelian inheritance. (a) What conclusion can you draw about the inheritance of these traits based on the \(\mathrm{F}_{1}\) results? (b) Which phenotype appears most frequently in the \(\mathrm{F}_{2}\) results? Write a mathematical expression that predicts the frequency of occurrence of this phenotype. (c) Which \(\mathrm{F}_{2}\) phenotype is expected to occur least frequently? Write a mathematical expression that predicts this frequency. (d) How often is either \(P_{1}\), phenotype likely to occur in the \(F_{2}\) generation? (e) If the \(F_{1}\) plant is testcrossed, how many different phenotypes will be produced?

A geneticist, in assessing data that fell into two phenotypic classes, observed values of \(250: 150 .\) He decided to perform chi- square analysis using two different null hypotheses: (a) the data fit a 3: 1 ratio; and (b) the data fit a 1: 1 ratio. Calculate the \(\chi^{2}\) values for each hypothesis. What can you conclude about each hypothesis?

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